Start learning 50% faster. Sign in now
Let E denote the event that the person selected is actually having COVID and A the event that the person's COVID test is diagnosed as +ive. We need to find P(E|A). Also, E’ denotes the event that the person selected is actually not having COVID. Clearly, {E, E'} is a partition of the sample space of all people in the population. We are given that P(E) = 0.1% = 0.1/100 = 0.001 P(E') = 1 – P(E) = 0.999 P(A|E) = P(Person tested as COVID+ive given that he/she is actually having COVID) = 90% = 90/100 = 0.9 and P(A|E') = P(Person tested as COVID +ive given that he/she is actually not having COVID) = 1% = 1/100 = 0.01 Now, by Bayes' theorem P(E|A) = [P(E) × P(A|E)]/[P(E) × P(A|E) + P(E') × P(A|E')] = [0.001 × 0.9]/[0.001 × 0.9 + 0.999 × 0.01] = 90/1089 = 0.083 approx.
Which of the following may be the code for ‘hope live’ ?
'F#P22' is the code of which of the following?
If ‘HAMPER’ is coded as '22', ‘SURFACE’ is coded as '58', then what is the code for ‘PAPER’ in the same code language?
In a certain code language, ‘DEMONS’ is written as ‘YSSPGE’. How will ‘GRAPES’ be written in that code language?